Documentation:CHBE Exam Wiki/Module 6 - Reactive energy Balances

CHBE 241 Exam resources wiki
Chemical and Biological Engineering
Welcome to the CHBE Exam Resources Wiki! This wiki is intended to host past exams with fully worked-out hints and solutions
Past Exams
Final Exam 2016W
Midterm Exam 1 2016W
Midterm Exam 2 2016W
Problem Sets
Module 1 - Process Basics
Module 2 - Reactors
Module 3 - Separations 1
Module 4 - Separations 2
Module 5 - Non-reactive Energy Balances
Module 6 - Reactive Energy Balances

Question 1

Trichloroethylene is produced in a two-step reaction sequence as shown in the steps below.

Reaction 1: C₂H₄(g) 2Cl₂ (g) → C₂H₂Cl₄(l) + H₂(g) where, ${\hat {H}}_{r}^{\circ }=-385.76kJ/mol$
Reaction 2: C₂H₂Cl₄(l) → C₂HCl₃(l) + HCl(g)

The standard heat of formation of tricholoroethylene (liquid) is - 276 2 kJ/mol.

Question 1a

Calculate the standard heat of the second reaction.

Solution

We first need to determine the heat of formation of tetrachloroethane

${\begin{aligned}({\hat {H}}_{f}^{\circ })_{C_{2}H_{2}Cl_{4}(l)}&=({\hat {H}}_{r1}^{\circ })+({\hat {H}}_{f}^{\circ })_{C_{2}H_{4}(g)}\\&=-333.48kJ/mol\\({\hat {H}}_{r2}^{\circ })&=({\hat {H}}_{f}^{\circ })_{C_{2}HCl_{3}(l)}+({\hat {H}}_{f}^{\circ })_{HCl(g)}-({\hat {H}}_{f}^{\circ })_{C_{2}H_{2}Cl_{4}(l)}\\&=-35.03kJ/mol\\\end{aligned}}$

Question 1b

Use Hess’s law to calculate the standard heat of the reaction
C₂H₄(g)+ 2Cl₂ (g) → C₂HCl₃(l) + H₂(g) + HCl(g)

Solution

Reaction (1) + Reaction (2)
${\begin{aligned}({\hat {H}}_{r}^{\circ })&=({\hat {H}}_{r1}^{\circ })+({\hat {H}}_{r2}^{\circ })\\&=-385.76+(-35.03)\\&=-420.79kJ/mol\\\end{aligned}}$

Question 1c

If 450 mol/h of C₂HCl₃ (l) is produced in the reaction of part (b) and the reactants and products are all at 25 C and 1 atm, how much heat is evolved or absorbed in the process?

Solution

Assume that
${\begin{aligned}{\dot {Q}}&=\Delta {\dot {H}}\\&=450mol/h\cdot (-420.79kJ/mol)\\&=-1.89\cdot 10^{5}kJ/h\\&=-52.60kW({\text{heat is evolved to the surroundings}})\\\end{aligned}}$

Question 2

Ethylene oxide is produced by the oxidation of ethylene under the following reaction:
C₂H₄(g) + 1/2 O₂ (g) → C₂H₄O(g)

Another competing reaction results in oxidation of ethylene to CO.
In a plant, the feed to the reactor contains 2 mol C₂H₄/ mol O₂ while the yield and the conversion in the reactor are 0.8 mol C₂H₄O produced/ mol C₂H₄ consumed and 40% respectively. The outlet stream from the reactor then separates into the following streams:

• C₂H₄ and O₂ are recycled back to the reactor
• C₂H₄O is sold
• CO₂ and H₂O are disposed
  

Also, the inlet and outlet streams from the reactor are kept at 450°C, while the feed stream and the streams leaving the separator are at 25°C.

Question 2a

Using a basis of 4 moles of ethylene entering the reactor, draw a process flow diagram for the plant.

Solution

Question 2b

Determine the flow rates (molar) and compositions of all the streams.

Solution

With 40% conversion, 1.6 mol of C₂H₄ is consumed. Therefore,
$n_{3}=2.4mol$
With 80% yield,
${\begin{aligned}n_{5}&=(1.6molC_{2}H_{4}consumed)\cdot 0.8\\&=1.28mol\\\end{aligned}}$

C balance on reactor:
${\begin{aligned}2(4)&=2(2.4)+2(1.28)+\cdot n_{6}\\n_{6}&=0.64mol\\\end{aligned}}$

Since the molar ratio of water to CO₂ produced is 1:1,
$n_{7}=0.64mol$

O balance on the reactor:
${\begin{aligned}2(2)&=2(n_{4})+1.28+2(0.64)+0.64\\n_{4}&=0.4mol\\\end{aligned}}$

Overall C balance:
${\begin{aligned}2n_{1}&=0.64+2(1.28)\\n_{1}&=1.6mol\\\end{aligned}}$

Overall O balance:
${\begin{aligned}2n_{2}&=2(0.64)+0.64+1.28\\n_{2}&=1.6mol\\\end{aligned}}$

Compositions:
Feed stream: 50% C₂H₄, 50% O₂
Recycle stream: 85.7% C₂H₄,14.2% O₂
Reactor inlet: 66.7% C₂H₄, 33.3% O₂
Reactor outlet: 44.8% C₂H₄, 7.5% O₂, 23.9% C₂H₄O, 11.9% CO₂, 11.9% H₂O

Question 2c

Determine the heat required for the whole process and the reactor alone respectively. Use the following information for C₂H₄O,
$({\hat {H}}_{f}^{\circ })=-51kJ/mol$
$C_{p}[J/(mol\cdot K)]=-4.69+0.2061T-9.995\cdot 10^{-5}T^{2}$ where T is in kelvin

Solution

By performing energy balance on the process:
${\begin{aligned}Q_{process}&=\Delta H\\&=H_{out}-H_{in}\\&=-583.7kJ\\Q_{reactor}&=-558.7kJ\\\end{aligned}}$

Question 2d

Determine the flow rate (kg/h) and the energy consumption in kW for a production for 5000kg C₂H₄O/day.

Solution

Mass of C₂H₄O produced with initial basis,
${\begin{aligned}m&=1.28mol\cdot 440.5kg/10^{3}mol\\&=0.0564kg\\{\text{Scale factor}}&={\frac {5000kg/day}{0.0564kg}}\\&=88677/day\\\end{aligned}}$

The feed mass at the initial basis,
${\begin{aligned}m_{feed}'&=1.6mol\cdot (28.05gC_{2}H_{4}/mol)+1.6mol\cdot (32gO_{2}/mol)\\&=96.08g\\\end{aligned}}$

The fresh feed rate for the production basis of 5000kg/day,
${\begin{aligned}{\dot {m}}_{production}&=96.08\cdot 10^{-3}kg\cdot 88677/day\\&=8520kg/day\\\end{aligned}}$

By performing energy balance on the process:
${\begin{aligned}{\dot {Q}}_{process}&=-583.7kJ\cdot 88677/day\cdot {\frac {1day}{24hr}}\cdot {\frac {1hr}{3600s}}\cdot {\frac {1kW}{1kJ/s}}\\&=-573.4kW\\{\dot {Q}}_{reactor}&=-599.1kW\\\end{aligned}}$